Question

我有问题。我创建了此类，以查找六边形的所有共享边：

public class HexagonRegistryList
{
    public int HexagonNum { get; set; }
    public float x1 { get; set; }
    public float y1 { get; set; }
    public float x2 { get; set; }
    public float y2 { get; set; }
    public float x3 { get; set; }
    public float y3 { get; set; }
    public float x4 { get; set; }
    public float y4 { get; set; }
    public float x5 { get; set; }
    public float y5 { get; set; }
    public float x6 { get; set; }
    public float y6 { get; set; }
    public int ShapeNum { get; set; }

    public HexagonRegistryList()
    {
        this.AdjacentShapeNumbers = new List<int>();
    }

    public List<int> AdjacentShapeNumbers { get; set; }

    public IEnumerable<(float x, float y)> GetPoints()
    {
        yield return (x1, y1);
        yield return (x2, y2);
        yield return (x3, y3);
        yield return (x4, y4);
        yield return (x5, y5);
        yield return (x6, y6);
    }


    public bool IsAdjacentTo(HexagonRegistryList other)
    {
        var isAdjacentTo =
                    GetPoints().Intersect(other.GetPoints()).Count() >= 2;
        if (isAdjacentTo)
        {
            if (other.ShapeNum != 0)
            {
                AdjacentShapeNumbers.Add(other.ShapeNum);
            }
        }
        return isAdjacentTo;
    }
}

但是现在我想要一些东西，因此值不必完全相同，但是它们可以具有最大1的差。因此，当我将350与350进行比较时，也可能是349与350的关系，或者350和351。有人可以帮我吗？

Answer 1

定义自定义比较器：

public struct PointComparer : IEqualityComparer<(float x, float y)>
{
    public bool Equals((float x, float y) p1, (float x, float y) p2)
    {
        return Math.Abs(p1.x - p2.x) < 1f && Math.Abs(p1.y - p2.y) < 1f;
    }

    public int GetHashCode((float x, float y) obj)
    {
        return 1;
    }
}

然后将其传递给Intersect方法

GetPoints().Intersect(other.GetPoints(), new PointComparer())

Answer 2

您将需要创建一个实现，该实现使用有限的绝对差来确定邻近度；可以按照以下步骤进行。

var DIFF_THRESHOLD = 1.0f;
var DIFF_THRESHOLD_SQUARE = DIFF_THRESHOLD * DIFF_THRESHOLD;

private IsCloseTo(float x1, float y1, float x2, float y2)
{
    var EuklideanDistance
        = ( x1 - x2 ) * ( x1 - x2 ) +  ( y1 - y2 ) * ( y1 - y2 );
    return EuklidenDistance <= DIFF_THRESHOLD_SQUARE;
}

然后，您可以使用Linq来确定彼此接近的点数。

找到相交点不正确的

2 个答案: